A076840 a(1) = a(2) = 1; a(n) = (a(n-1) + 1)/a(n-2) (for n>2, n odd), (a(n-1)^2 + 1)/a(n-2) (for n>2, n even).

1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2, 5, 3, 2, 1, 1, 2

Offset

1,3

Comments

Any sequence a(1),a(2),a(3),... defined by the recurrence a(n) = (a(n-1)+1)/a(n-2) (for n>2, n odd), (a(n-1)^2+1)/a(n-2) (for n>2, n even) has period 6. The theory of cluster algebras currently being developed by Fomin and Zelevinsky gives a context for these facts, but it doesn't really explain them in an elementary way. - James Propp, Nov 20 2002

Links

Table of n, a(n) for n=1..105.

Sergey Fomin and Andrei Zelevinsky, Cluster algebras II: Finite type classification, arXiv:math/0208229 [math.RA], 2002-2003.

Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Maple

a := 1; b := 1; f := proc(n) option remember; global a, b; if n=1 then RETURN(a); fi; if n=2 then RETURN(b); fi; if n mod 2 = 1 then RETURN((f(n-1)+1)/f(n-2)); fi; RETURN((f(n-1)^2+1)/f(n-2)); end;

Mathematica

LinearRecurrence[{0, 0, 0, 0, 0, 1}, {1, 1, 2, 5, 3, 2}, 105] (* Jean-François Alcover, Nov 22 2017 *)

Crossrefs

Cf. A076839, A076841, A076844.

Sequence in context: A242292 A159897 A019709 * A328823 A364143 A346265

Adjacent sequences: A076837 A076838 A076839 * A076841 A076842 A076843

Keyword

nonn,easy

Author

N. J. A. Sloane, Nov 21 2002

Status

approved